Why does entailment not hold in this case?

Question

I am having issue understanding something.

For the first formula:

$$(\exists z (R(z) \implies Q(z)))$$ This makes sense that this is true, because in interpretation $I$, there exists a $z$ ($z = 2$) such that $R(z)$ is false, thus $\exists z (R(z) \implies Q(z))$ must hold true vacously.

But how is

$$(\exists z R(Z)) \implies (\exists z Q(z))$$ A false statement?

If we take $z = 2$ then $\exists z R(z)$ is false thus this implication must be true

Can someone explain? Thanks very much!

Answer 1

If you take instead $z=1,$ then $\exists z R(z)$ is true while $\exists z Q(z)$ is false, provided you mean in your interpretations that $R(1)$ is true but $Q(z)$ is always false for any $z.$

Answer 2

If you take $z=2$, then $Rz$ is false. Not $\exists z:Rz$, just $Rz$.

However if you take $z=1$, then $Rz$ is true.

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But we are not talking about the statement $Rz$. We are talking about

$$\exists z: Rz$$

which is true if $Rz$ is true for at least one $z$. And since $Rz$ is true for $z=1$, the statment $\exists z:Rz$ is true.